'Matchless' printed from http://nrich.maths.org/

Listen to Jenny and Graeme talking together about the problem. [ audio ]

Encounters with simultaneous equations can become over-familiar, routine experiences for students. This type of problem causes a "stop and think" moment, requires some problem-solving ingenuity, and leads into a consideration of redundancy of information. Students might make a start by substituting some arbitrary $x$, $y$ values to get a feel for the problem and to grasp that the five expressions
don't generally take the same value.

This is in contrast to expressions that are identities, for example $2 (x+y) - 3(x-y )$ and $5y - x$, where the two expressions take the same value for any $x$, $y$ combination. This idea is worth some discussion.

Questions or prompts:

For a start:

$2x + 3y - 20$ equals $ 5x - 2y +38$ ...

Could you find an $x$, $y$ pair that works for two, for three, or for four of the expressions but not for all of them ?

Further ideas:

Make up a similar problem of your own.Or extending that : can you create a similar problem with an "odd one out"? That is, one expression which does not equal the other four, which are equal for some specific $x$, $y$ pair.

The following interesting account was sent in by a class teacher working withYear 8s in Maths Club at St Albans High School for Girls

Becky worked as follows:

$2x + 3y -20 = 4x + 5y -72$

($-2x$ to each side)

$3y -20 = 2x + 5y - 72$

($+72$ to each side)

$3y + 52 = 2x + 5y$

($-3y$ to each side)

$52 = 2x + 2y$

Then, $5x - 2y + 38 = x - 4y + 108$

($-x$ from each side)

$4x - 2y + 38 = -4y + 108$

($+4y$ to each side)

$4x + 2y + 38 = 108$

( $-38$ to each side)

$4x + 2y = 70$

Becky looked at the difference between these two equations and deduced

$2x = 18, so x = 9$

Then using one of her equations, and substituting $x = 9$ she found $y = 17$.

Ele and Sarah reached the same conclusion but started out by looking at

all the possible pairs of expressions. Then they selected the following two

as easiest to work with:

From $2x + 3y - 20 = 4x + 5y - 72$

They deduced $26 = x + y$

From $2x + 3y - 20 = x - 4y + 108$

They deduced $128 = x + 7y$

From these two equations they deduced $6y = 128 - 26$

$6y = 102$

$y= 17$

Then from $x + y = 26$ they found $x = 9$

All three girls were then challenged to decide how much of the information

they needed to use to solve the problem. Their conclusion was only $3$

statements were needed; they could have used $A = B$ and $A = C$ to deduce the answer where $A,B,C$ label different expressions given.